I am to find a holomorphic function on $\mathbb{C}\setminus\{0\}$ with $u(x,y)=\dfrac{x+y}{x^2+y^2}$ and $f(1)=1$. Preceding this, I had to show we could write $f'(z) = \frac{\partial u}{\partial x} -i\frac{\partial u}{\partial y}$, which is clearly supposed to be a hint for this part.
The issue is I don't see how it helps at all. Other than taking $u(x,y)$ as I know it and doing lots of differentiation/integration along with the Cauchy Riemann equations (which would be a lot of work..) the only idea I had that used this as a hint was to compute $f'(z)$ with that formula, and see if it would arrange into an expression in $z$ which was readily integrable. I got to $f'(z(x,y)) = \dfrac{(y+ix)^2+i(y+ix)^2}{(x^2+y^2)^2}$, but given this all assumes $x$ is the real part, etc., you can't just swap $x$ and $y$ and there's no way to get to $x+iy$ nicely! so that approach seemed to fail..
This is a question from a past exam paper; one of the short 'more routine' questions in the first section. Typically these aren't so bad, but with this particular question is being more difficult to deal with.
Best Answer
Since $u(x,y) = \dfrac{x+y}{x^2+y^2}$,
$$\frac{\partial u}{\partial x} = \frac{1}{x^2+y^2}-\frac{2x(x+y)}{(x^2+y^2)^2} = \frac{x^2+y^2-2x^2-2xy}{(x^2+y^2)^2} = \frac{-x^2-2xy+y^2}{(x^2+y^2)^2}.$$
Likewise, by symmetry,
$$\frac{\partial u}{\partial y} = \frac{-y^2-2xy+x^2}{(x^2+y^2)^2}.$$
Putting these together shows that
$$\begin{align} f'(z) &= \frac{-x^2-2xy+y^2-i(-y^2-2xy+x^2)}{(x^2+y^2)^2} \\ &= \frac{-x^2-2xy+y^2+iy^2+2ixy-ix^2}{(x^2+y^2)^2}.\end{align}$$
The numerator can be rewritten a bit:
$$\begin{align}(-x^2+2ixy+y^2)+i(-x^2+2ixy+y^2) &= (1+i)(-x^2+2ixy+y^2) \\ &= -(1+i)(x^2-2ixy-y^2) \\ &= -(1+i)(x-iy)^2.\end{align}$$
Thus we have that
$$f'(z) = \frac{-(1+i)(x-iy)^2}{((x+iy)(x-iy))^2} = -\frac{1+i}{(x+iy)^2} = -\frac{1+i}{z^2}.$$
From here, it is a simple application of integration to get the final answer.